A tank 9 m long, 4 m wide, and $$\frac{1}{3}$$​ m deep is dug in a field 24 m long and 14 m wide. If the dug soil is spread evenly over the rest of the field, then by how much will the height of the field increase?

Correct option is D

Given:

Dimensions of tank (dug portion):
Length = 9 m, Width = 4 m, Depth = $$\frac{1}{3}$$​ m
Dimensions of field:
Length = 24 m, Width = 14 m

Formula Used:

The volume of cuboid = Length $$\times$$​ Breadth $$\times$$​ Height
Area of Rectangle = Length $$\times$$​ Breadth
Solution:

Volume of tank = Length × Width × Depth

Volume = $$9 \times 4 \times \frac{1}{3}$$

​​Volume = 12 cubic meters

The total field area is:

Total Area = 24 $$\times$$​ 14 = 336 square meters

Tank Area = 9 $$\times$$​ 4 = 36 square meters

Remaining Area = 336 - 36 = 300 square meters

The soil from the tank is spread over the remaining 300 m²:

Increase in Height = $$\frac{\text{Volume of Dug Soil}}{\text{Remaining Area}}$$

Increase in Height = $$\frac{12}{300} = 0.04m=4cm$$

So, the height of the field will increase by 4 cm.

Thus, the correct answer is (d).